A person walks 15.0 m in 5.00 s and then
walks 12.0 m in 10.00 s. What is the
average speed of the person?
Answer:
Explanation:
Given:
D₁ = 15.0 m
t₁ = 5.00 s
D₂ = 12.0 m
t₂ = 10.00 s
___________
V - ?
The average speed of the person:
V =(D₁ +D₂) / (t₁ + t₂)
V =(15.0 + 12.0) / (5.00 + 10.00) = 27.0 / 15.00 ≈ 1.8 m/s
A small electric motor produces a force of 3 N that moves a remote-control car 4 m every second. How much power does the motor produce?
f = force = 5N
v = velocity = 4m/s
p = power
p = force * velocity
Replacing:
p= 3N * 4 m/s = 12W
The motor produces 12W of power.
I have this homework i need help with ?Part (a)
Given that the mass of train is
[tex]m=\text{ }8.12\times10^6\text{ kg}[/tex]The deceleration of the train is
[tex]\begin{gathered} a\text{ = -104 km/h} \\ =-104\times\frac{5}{18} \\ =\text{ -28.88 m/s} \end{gathered}[/tex]The braking force can be calculated by the formula
[tex]F=ma[/tex]Substituting the values, the braking force will be
[tex]\begin{gathered} F=8.12\times10^6\times(-28.88) \\ =2.348\times10^8\text{ N} \end{gathered}[/tex]Thus, the braking force is 2.348 x 10^8 N.
Part 1)If a force of magnitude 125 N is exerted horizontally as shown, find the force exerted by the hammer claws on the nail. (Assume that the force the hammer exerts on the nail is parallel to the nail).Answer in units of N.Part 2)Find the force exerted by the surface on the point of contact with the hammer head. Assume that the force the hammer exerts on the nail is parallel to the nail. Answer in units of N.
Given data:
* The force acting on the hammer in the horizontal direction is F = 125 N.
* The distance of horizontal force from the point of contact is d = 28 cm.
* The angle of the nail with the vertical direction is,
[tex]\theta=26^{^{\circ}}[/tex]* The distance of the nail from the point of contact is x = 5.86 cm.
Solution:
The moment caused by the horizontal force on the hammer is equal to the moment caused by the horizontal component of force at the nail about the point of contact, thus,
[tex]F\times d=F^{\prime}x\text{ sin(}\theta)[/tex]Substituting the known values,
[tex]\begin{gathered} 125\times28=F^{\prime}\times5.86\times\sin (26^{\circ}) \\ 3500=F^{\prime}\times2.57 \\ F^{\prime}=\frac{3500}{2.57} \\ F^{\prime}=1361.87\text{ N} \end{gathered}[/tex]Thus, the force exerted on the nail is 1361.87 N or approximately 1362 N.
determine the total force and absolute pressure on the bottom of a swimming pool 28m by 8.5m whose uniform depth is 1.8. what will be the pressure against the side of the pool near the bottom
Answer:
Explanation:
Let us draw the pool for our visualization:
Pressure is defined as Force / area.
Now, what is the total force on the bottom of the pool?
It will be equal to the force due to the atmospheric pressure + force due to the mass of the water.
If we call p the density of water, then the force is
[tex]F=mg=\text{pVg}[/tex]where V = volume of the pool.
Therefore, the pressure due to water is
[tex]P=\frac{\rho Vg}{A}[/tex]Here remember that V = abh and A = ab; therefore, the above gives
[tex]P=\frac{\rho(abh)g}{ab}[/tex][tex]\boxed{P=\rho gh\text{.}}[/tex]adding the atmospheric pressure gives
[tex]\boxed{P=\rho gh+P_{\text{atm}}}[/tex]This is the pressure exerted on the bottom of the pool. But what about the sides?
Now near the bottom of the pool, the pressure exerted on the walls will be about the same as that exerted on the bottom of the pool. Therefore, we can use the above equation to find the pressure on the walls near the bottom.
SInce p = 1000kg / m^3, g = 9.8 m s^2, P_atm = 101325 Pascals, and h = 1.8 m, the equation gives
[tex]P=(\frac{1000\operatorname{kg}}{m^3})\times(\frac{9.8m}{s^2})\times(1.8m)+101325(N/m)[/tex]which evaluates to give
[tex]P=118965(N/m)[/tex]or in scientific notation, the above is
[tex]\boxed{P=1.19\times10^5(N/m)}[/tex]which is our answer!
Note that we also added to atmospheric pressure. This pressure will of course be canceled with the outside atmospheric pressure, leaving only the pressure due to water in effect.
what is the mass of a 2000 pound elephant in kg?
ANSWER:
907.2 kilograms
STEP-BY-STEP EXPLANATION:
To convert pounds to kilograms we must resort to the conversion factor between both units.
Where 1 pound is equal to 0.4536 kilograms.
Therefore:
[tex]2000\text{ pounds}\cdot\frac{0.4536\text{ kg}}{1\text{ pound}}=907.2\text{ kg}[/tex]The mass of the elephant is 907.2 kilograms
what force does the ground exert on a 1,000 kg car as it moves at 15 m/s through a dip in the road with a path radius of 30 m?
Answer:
Science; Physics; Physics questions and answers; 3. What force does the ground exert on a 1,000 kg car as it moves at 15 m/s through a dip in the road with a path radius of 30 m.
Explanation:
Glade 2 help :)
A cube 6.0 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.5 cm in diameter all the way through and perpendicular to one face, you find that the cube weighs 6.30 N. 1. What is the density of the metal? 2. What did the cube weigh before you drilled the hole in it?
1.
First, let's calculate the volume of the cube:
[tex]V_1=6^3=216\text{ cm^^b3}[/tex]Now, let's calculate the volume of the cylinder drilled:
[tex]\begin{gathered} V_2=\frac{\pi d^2h}{4}\\ \\ V_2=\frac{3.14159\cdot2.5^2\cdot6}{4}\\ \\ V_2=29.45\text{ cm^^b3} \end{gathered}[/tex]So the volume of the cube after being drilled is:
[tex]\begin{gathered} V=V_1-V_2\\ \\ V=216-29.45\\ \\ V=186.55\text{ cm^^b3} \end{gathered}[/tex]If the weight is 6.3 N, let's find the mass:
[tex]\begin{gathered} W=m\cdot g\\ \\ 6.3=m\cdot9.8\\ \\ m=\frac{6.3}{9.8}\\ \\ m=0.643\text{ kg} \end{gathered}[/tex]And the density of the metal is:
[tex]\begin{gathered} d=\frac{m}{V}\\ \\ d=\frac{0.643}{186.55}\\ \\ d=0.0034468\text{ kg/cm^^b3}\\ \\ d=3.4468\text{ g/cm^^b3} \end{gathered}[/tex]2.
To find the weight of the cube before being drilled, let's use the following rule of three:
[tex]\begin{gathered} volume\rightarrow weight\\ \\ 216\text{ cm^^b3}\rightarrow x\text{ N}\\ \\ 186.55\text{ cm^^b3}\rightarrow6.3\text{ N}\\ \\ \\ \\ \frac{216}{186.55}=\frac{x}{6.3}\\ \\ x=\frac{216\cdot6.3}{186.55}\\ \\ x=7.295\text{ N} \end{gathered}[/tex]Suppose your on the bus and walking from the back to the front while the bus is moving at 10 mph. If you walk at 2 mph, what is the speed in mph?
Given data:
* The speed of the bus is 10 mph.
* The speed of the person in the bus is 2 mph.
Solution:
As the lecture hall is in state of rest with respect to Bus,
Thus, the speed of the person walking inside the bus with respect to the lecture hall is,
[tex]\begin{gathered} S=\text{speed of bus+spe}ed\text{ during walk} \\ S=10\text{ mph + 2 mph} \\ S=12\text{ mph} \end{gathered}[/tex]Thus, the speed relative to the lecture hall is 12 mph.
True or False: Wave-particle duality describes how some small particles act like both waves and particles.
Answer:
False
Explanation:
The wave-particle duality says that every particle can act sometimes like a wave and sometimes like a particle. This applies to all particles, so the statement: wave-particle duality describes how some small particles act as both waves and particles is False.
A baseball is hit from theground straight up into the airwith a speed of 14.6 m/s. Howlong is the ball in air (time inseconds from ground-to-ground)?
Givens.
• Initial speed = 14.6 m/s.
,• FInal speed = 0 m/s (at highest point).
,• Gravity = 9.8 m/s^2.
First, find the time needed to reach the highest point.
[tex]\begin{gathered} v_f=v_0+gt \\ t=\frac{v_f-v_0}{g} \\ t=\frac{0-14.6\cdot\frac{m}{s}}{-9.8\cdot\frac{m}{s^2}} \\ t\approx1.49\sec \end{gathered}[/tex]It takes 1.49 seconds to reach the highest point.
The time that the baseball takes to reach the ground is double t because the trajectory is symmetrical, that is, it takes the same time to go from ground level to highest point than from highest point to ground level.
[tex]t_{\text{total}}=2\cdot1.49\sec =2.98\sec [/tex]Therefore, the baseball is 2.98 seconds in the air.
What is the height intercept?(b) What is the slope of the line?185.5 cm/hours(c) What is the height intercept?(d) Assuming Laura started with an empty pool at time = 0 hours, you would expect the height intercept to be zero. Unfortunately due to error in measurements theintercept was not zero. Use the 5% Rule to calculate the height intercept error. What is the the height intercept error?(e) What would you expect the height of the water to be after 14 hours?
C) the height intercept would be 0 given the pool has no water at the beggining.
D) To find the vertical axis intercept error, we will need to use the formula
[tex]Error=\lvert{\frac{vertical\text{ }Axis}{Largest\text{ }Value}}\rvert *100[/tex]A car moves 25km north and then another 25km west after that it moves 40km north and then 20km west. Find the displacement
Given:
The distance traveled by car;
d₁=25 km to the north
d₂=25 km to the west.
d₃=40 km to the north.
d₄=20 km to the west.
To find:
The displacement of the car.
Explanation:
The displacement of an object is the shortest distance between its initial and final position.
Referring to the diagram, D is the displacement of the car.
The total displacement of the car is given by,
[tex]D=\sqrt{(d_1+d_3)^2+(d_2+d_4)^2}[/tex]On substituting the known values,
[tex]\begin{gathered} D=\sqrt{(25+40)^2+(25+20)^2} \\ =79.06\text{ m} \end{gathered}[/tex]Final answer:
Thus the total displacement of the car is 79.06 m
An athlete starts at point A and runs at a constant speed of around a circular track 100 m in diameter
, as shown in Fig. P3.40 below. Find the x and y-components of this runner’s average velocity and average acceleration between points
(a) A and B, (b) A and C, (c) C and D, and (d) A and A (a full lap). (e) Calculate the magnitude of the runner’s average velocity between A and B. Is his average speed equal to the magnitude of his average velocity? Why or why not? (f) How can his velocity be changing if he is running at constant speed?
a ) The x and y-components of average velocity and average acceleration between points A and B are 3.8 m/s, 3.8 m/s and 0.46 m/s², - 0.46 m/s²
e ) The magnitude of the runner’s average velocity between A and B is
t = 2 π r / v
t = 2 * 3.14 * 50 / 6
t = 52.4 s for full lap
t per quarter = 52.4 / 4 = 13.1 s
v = Δx / Δt
a = Δv / Δt
a ) From A to B,
vx = ( 0 - ( - 50 ) ) / 13.1
vx = 3.8 m / s
vy = ( 50 - 0 ) / 13.1
vy = 3.8 m / s
ax = ( 6 - 0 ) / 13.1
ax = 0.46 m / s²
ay = ( 0 - 6 ) / 13.1
ay = - 0.46 m / s²
b ) From A to C,
t = 52.4 / 2
t = 26.2 s
vx = ( 50 - ( - 50 ) ) / 26.2
vx = 3.8 m / s
vy = 0
ax = 0
ay = ( - 6 - 6 ) / 26.2
ay = - 0.46 m / s²
c ) From C to D,
t = 13.1 s
vx = ( 0 - 50 ) / 13.1
vx = - 3.8 m / s
vy = ( - 50 - 0 ) / 13.1
vy = - 3.8 m / s
ax = ( - 6 - 0 ) / 13.1
ax = - 0.46 m / s²
ay = ( 0 - ( - 6 ) ) / 13.1
ay = 0.46 m / s²
d ) From A to A,
Since the starting and ending points are exactly the same, there is no displacement. So the average velocity will be zero. Due to no change in velocity, there will be no acceleration
e ) From A to B,
v = √ vx² + vy²
v = √ 3.8² + 3.8²
v = 5.4 m / s
Displacement is the shortest distance between two points. So it will basically be a straight line. But the athlete runs in a circular motion. So distance will be larger than the displacement. So speed will be higher than velocity.
s = 6 m / s
v = 5.4 m / s
s > v
f ) At constant speed in a circular motion, only the magnitude is constant. Its direction keeps changing. So velocity cannot be constant in a circular motion.
Therefore,
a ) vx = 3.8 m / s, vy = 3.8 m / s ; ax = 0.46 m / s², ay = 0.46 m / s²
b ) vx = 3.8 m / s, vy = 0 ; ax = 0, ay = - 0.46 m / s²
c ) vx = - 3.8 m / s, vy = - 3.8 m / s ; ax = - 0.46 m / s², ay = 0.46 m / s²
d ) vx = 0, vy = 0 ; ax = 0, ay = 0
e ) v = 5.4 m / s
f ) Due to change in direction.
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A 6.00 kg pendulum bob is placed on a 10.0 m long string and pulled back 5.00°. What is the period of the pendulum when it is released?3.84 s8.03 s6.35 s1.27 s
As we know that time period for a pendulum is,
[tex]\begin{gathered} T=2\pi\sqrt[\placeholder{⬚}]{\frac{l}{g}} \\ Here, \\ l=10m \\ g=9.8\text{ m/s}^2 \\ So, \\ T=2\times3.14\sqrt[\placeholder{⬚}]{\frac{10}{9.8}}=6.343s \\ \end{gathered}[/tex]So 3rd option is correct option.
7.3 kg of copper sits at a temperature of 38 degrees F. How much heat is required to raise its temperature to 865 degrees F? The specific heat of copper is 385 J/kg- degree C. Submit your anwser in exponential form.
The heat Q needed to increase the temperature of a sample with mass m and specific heat c by an amount ΔT is:
[tex]Q=mc\Delta T[/tex]On the other hand, a change in temperature in Farenheit is related to a change in temperature in Celsius as:
[tex]\Delta T_C=\frac{5ºC}{9ºF}\Delta T_F[/tex]Replace m=7.3kg, c=385J/(kgºC), as well as the final and initial temperatures to find the heat required to raise the temperature of the sample of Copper:
[tex]Q=(7.3\operatorname{kg})(385\frac{J}{\operatorname{kg}ºC})(865ºF-38ºF)[/tex]Since the specific heat is given in units of Joules per kilogram per degree Celsius, introduce the factor 5ºC/9ºF to write the change in temperature in degrees Celsius:
[tex]\begin{gathered} Q=(7.3\operatorname{kg})(385\frac{J}{\operatorname{kg}ºC})(865ºF-38ºF)\times\frac{5ºC}{9ºF} \\ =1,291,268.611\ldots J \\ \approx1.3\times10^6J \end{gathered}[/tex]Therefore, the amount of heat required to raise the temperature of the 7.3 kg of Copper sample from 38ºF to 865ºF, is 1.3*10^6 Joules.
As a sports psychologist how would you suggest working with an athlete who suffers from self-
attention in front of the home crowd?
As a sports psychologist my suggestion would be to ensure that they maintain a balanced approach about others opinions and should concentrate on their performance than the opinion of other.
Many athletes are intimidated by or motivated by the support and encouragement of the home audience when competing in front of them. They might make an effort to get people's respect and favor. Although it can improve performance, it can also negatively increase an athlete's pressure to perform or even cause them to feel overconfident. Athletes could feel humiliated if they lose or fail because they couldn't satisfy the expectations of their audience. Additionally, they can lose self-respect, which might have an effect on their morale and future performance.
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what is the weight of a 500 g package of spaghetti noodles ? (A) in N (B) in oz
Given:
Weight of the spaghetti noddles = 500 grams
Let's find the weight of the noodles in Newtons(N) and Ounce(oz).
First convert 500g to kg:
[tex]500g=\frac{500}{1000}=\frac{1}{2}kg[/tex](A) in N
Where 1 kg = 9.8 N
Thus, we have:
[tex]\frac{1}{2}kg=\frac{1}{2}\ast9.8=4.9N[/tex]The weight of a 500 g package of spaghetti noodles in Newton is 4.9 N
(B) in oz
Where:
1 gram = 0.035274 ounce
Thus, we have:
[tex]500g=500\ast0.035274=17.637\text{ oz}[/tex]The weight of a 500g package of spaghetti noodles in Ounce is 17.637 oz
ANSWER:
(A) 4.9 N
(B) 17.637 oz
Block A hangs from a light string that passes over a light pulley and is attached to block B, which is on a level horizontal frictionless table as shown above. Students are to determine the mass of block B from the motion of the two-block system after it is released from rest. They plan to measure the time block A takes to reach the floor. The students must also take which of the following measurements to determine the mass of block B?A. Only the mass of block A.B. Only the mass of Block A falls to reach the floorC. Only the mass of block A and the distance block A falls to reach the floor D. The mass of block A, the distance block A falls to reach the floor, and the radius of the pulleyUsing symbols like mA, mB, g, and a, wrote an equation that expresses Newton’s Second Law on the entire system as a whole. (Hint: Fnet = m•a) Which letter in the equation you just wrote would need to be measured before you could solve for mB? Assume you already know the value of g. One of the letters you should have written was a. The students are going to measure the time it takes for block A to reach the floor. What other quantity needs to be measured to find the acceleration? What equation would need to be used to find the acceleration?
Required: the mass of the block B.
Explanation:
we assume that mass of the block A is
[tex]m_A[/tex]and mass of the block B is
[tex]m_B[/tex]now, look at the free body diagram
from the above diagram, we can apply newton's law. we assume that both the block moves with the same acceleration a.
for block A
[tex]m_Ag-T=m_Aa.....(1)[/tex]for block B
[tex]T=m_Ba[/tex]from equations 1 and 2
we can write
[tex]\begin{gathered} m_Ag-m_Ba=m_Aa \\ a=\frac{m_Ag}{m_A+m_B}........(2) \end{gathered}[/tex]this is the acceleration of the whole system.
By the above equation, we can calculate the acceleration of the both the blocks.
we are interested in determining the mass of block mB.
we assume that block mA moves with above acceleration.
we know that
[tex]h=ut+\frac{1}{2}at^2......(3)[/tex]if we know the height of the block A and measure the time to reach the block A to the ground.
by the equation 3 we can calculate the acceleration.
[tex]a=\frac{2h}{t^2}......(4)[/tex]from the equation 2 and 4, we can write
[tex]\begin{gathered} \frac{2h}{t^2}=\frac{m_{A}g}{m_{A}+m_{B}} \\ m_A+m_B=\frac{t^2m_Ag}{2h} \\ m_B=\frac{t^{2}m_{A}g}{2h}-m_A......(5) \end{gathered}[/tex]by the above equation, we can easily can calculate the mass of the block B.
(a) part
only the mass of the block A and the distance block A falls to reach the floor is correct answer.
As we can see from the above equation.
I’m not sure how to even begin this problem. I know it’s an equilibrium problem
m= 27 kg
angle = 40°
mg = 27 x 9.8 = 264.6 N
Fty . x1 - mg (x1/2 ) = 0
Fty =
I have a question of a test that I already took and just want to know why my answer was incorrect, so when I spoke to the prevuios tutor and gave hime the question wuth the possible answer he asked me if it was for a graded test , and I said that it was a graded test and he said to report me , my question is a tutor can't help with an answer of a test that I took and my answer was incorrect, and I just want to understand why is it incorrect
The correct answer is (D)
Strong nuclear forces are responsible for holding together the nucleus of an atom; weak nuclear forces are involved when certain types of atoms break down.
A 3.10-kg block is moving to the right at 2.60 m/s just before it strikes and sticks to a 1.00-kg block initially at rest. What is the total momentum of the two blocks after the collision? Enter a positive answer if the total momentum is toward right and a negative answer if the total momentum is toward left.
Given,
The mass of the block moving to the right, M=3.10 kg
The speed of the block, u=2.60 m/s
The mass of the block at rest, m=1.00 kg
The total momentum of the two-block system before the collision is given by
[tex]p_i=Mu_{}[/tex]On substituting the known values,
[tex]\begin{gathered} p_i=3.10\times2.60 \\ =8.06\text{ kg}\cdot\frac{m}{s} \end{gathered}[/tex]From the law of conservation of energy, the total momentum of a system always remains the same.
Thus the momentum after the collision is equal to the total momentum of the blocks before the collision.
Thus the total momentum of the two blocks after the collision is 8.06 kg·m/s
A coil of 17.771 H carries a current of 11.121 A. Compute the energy stored .
Given
The inductance is H=17.771 H
Current is , I=11.121A
To find
The energy stored
Explanation
The energy stored is given by
[tex]E=\frac{1}{2}LI^2[/tex]Putting the values,
[tex]\begin{gathered} E=\frac{1}{2}\times17.771\times(11.121)^2 \\ \Rightarrow E=1098.92J \end{gathered}[/tex]Conclusion
The energy stored is 1098.92J
I have a homework problem that I don’t even know where to start or what formulas to use
We will have the following:
a) The velocity after 5 seconds will be:
[tex]v_f=v_o+at[/tex][tex]v=2.79m/s+(9.8m/s^2)(5s)\Rightarrow v=51.79m/s[/tex]So, the velocity after 5 seconds will be 51.79m/s.
b) We will have that the distance below the helicopter will be:
[tex]d=\frac{1}{2}(v_0+v_f)t[/tex][tex]d=\frac{1}{2}(2.79m/s+51.79m/s)(5s)\Rightarrow d=136.45m[/tex]So, it will be 136.45 m below the helicopter.
c) We will have that the velocity and distance given that the helicopter is moving constantly at 2.79m/s will be:
[tex]v=-2.79m/s+(9.8m/s^2)(5s)^2\Rightarrow v=46.21m/s[/tex]So, the velocity would be 46.21 m/s.
[tex]d=\frac{1}{2}(-2.79m/s+46.21m/s)(5s)\Rightarrow d=108.55m[/tex]So, the distance would be 108.55 m below the helicopter.
A particle moving along the x axis has a position given by x = (24t – 2.0t 3) m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero?
The velocity is defined by:
[tex]v=\frac{dx}{dt}[/tex]where x is the position of the particle and t is the time.
Plugging the position function given we have that the velocity is:
[tex]\begin{gathered} v=\frac{dx}{dt} \\ =\frac{d}{dt}(24t-2t^3) \\ =24-6t^2 \end{gathered}[/tex]Hence the velocity is given by the function:
[tex]v=24-6t^2[/tex]to determine the isntant when the velocity is zero we equate its expression to zero and solve for t:
[tex]\begin{gathered} 24-6t^2=0 \\ 6t^2=24 \\ t^2=\frac{24}{6} \\ t^2=4 \\ t=\pm\sqrt[]{4} \\ t=\pm2 \end{gathered}[/tex]Since time is always positive we conclude that the velocity is zero at t=2 s.
Now that we know at which instant the velocity is zero we need to remember that the acceleration is defined as:
[tex]a=\frac{dv}{dt}[/tex]then we have that:
[tex]\begin{gathered} \frac{dv}{dt}=\frac{d}{dt}(24-6t^2) \\ =-12t \end{gathered}[/tex]hence the acceleration is:
[tex]a=-12t[/tex]Plugging the value we found for the time we have that:
[tex]a(2)=-12(2)=-24[/tex]Therefore the acceleration of the particle when its velocity is zero is -24 meters per second per second.
A 98 N person is standing on a board, 1 m from the end. The board is balanced on a point that is 2m from the same end. The board is 49 N. How long is the board overall?
ANSWER:
8 m
STEP-BY-STEP EXPLANATION:
To calculate the length is the board overall we must apply the center of mass formula, which is as follows:
[tex]x_{cm}=\frac{m_{\text{board}}\cdot x_{\text{noard}}+m_p\cdot x_p}{m_{\text{board}}+m_p}[/tex]The mass of the person and that of the board, we calculate it as follows:
[tex]\begin{gathered} F_{\text{p}}=m_{\text{p}}\cdot g_{} \\ m_{\text{p}}=\frac{F_{\text{p}}}{g}=\frac{98}{9.8}=10kg \\ F_{\text{board}}=m_{\text{board}}\cdot g_{} \\ m_{\text{board}}=\frac{F_{\text{board}}}{g}=\frac{49}{9.8}=5kg \end{gathered}[/tex]Replacing:
[tex]\begin{gathered} -2=\frac{5\cdot x_{\text{board}}+10\cdot(-1)}{5+10} \\ (-2)(15)+10=5\cdot x_{\text{board}} \\ x_{\text{board}}=\frac{-20}{5} \\ x_{\text{board}}=-4\text{ m} \end{gathered}[/tex]Since the CM of the board is only 4 m from the edge of the board, and the MC of the board is at its center, the board is 8 m long.
Rate at which an object with a direction?
Rate at which an object changes its direction is velocity.
So the answer is velocity.
Question 8 of 10Objects with the same charge will:O A. have no effect on each other.B. exert a pulling force on each other.O C. repel each other.O D. attract each other.SU
We will have that:
objects with the same charge will repel each other.
A hiker walks 14.91 m, N and 4.40 m, E. What is the magnitude of his resultant displacement?
Givens.
• 14.91 meters North.
,• 4.40 meters East.
First, make a diagram to visualize the vectors and the resultant displacement.
In the figure, the purple vector d represents the resultant displacement, which horizontal component is 4.40m and its vertical component is 14.91m.
Let's use the following formula to find the resultant.
[tex]d=\sqrt[]{(y_{})^2+(x)^2}[/tex]Where y = 14.91 and x = 4.40.
[tex]\begin{gathered} d=\sqrt[]{(14.91m)^2_{}+(4.40m)^2} \\ d=\sqrt[]{222.31m^2+19.36m^2} \\ d=\sqrt[]{241.67m^2} \\ d\approx15.55m \end{gathered}[/tex]Therefore, the magnitude of the resultant displacement is 15.55m.
But, the resultant displacement refers to the vector, which is the following
[tex]d=(4.4i+14.91j)m[/tex]A man attempts to move a truck by pushing it, but he can't move it. Describe the work done by the man.
If a man tried to move a truck by pushing it, but he is not able to move it, then the work done by the man will be equal to zero.
What is Work?In physics, the word "work" involves measurement of energy transfer that takes place when an item is moved over a range by an externally applied, at least a portion of which is applied with in the direction of the displacement. The length of the path is multiplied by the element of a force acting all along the path to calculate work if the force is constant. The work W is theoretically equivalent towards the force f times the length d, or W = fd, to portray this concept.
As per the given question,
The man tries to move the truck, but he is not able to move. It means that the total displacement is zero, which means according to the formula of work done the work is also zero.
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